A semi-circular current-carrying wire having | vedclass

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(B) The magnetic force on a small current element $I d\vec l$ is given by $d\vec F = I (d\vec l 	imes \vec B)$.For a semi-circular wire in the $x-y$

Vedclass · Accepted Answer

$+ y-	ext{axis}$

Vedclass · Answer

(B) The magnetic force on a small current element $I d\vec l$ is given by $d\vec F = I (d\vec l 	imes \vec B)$.For a semi-circular wire in the $x-y$ plane, a small element at angle $	heta$ (measured from the negative $x$-axis) is $d\vec l = R d	heta (\cos 	heta \hat i + \sin 	heta \hat j)$.The magnetic field at this point is $\vec B = \frac{B_o x}{2R} \hat k = \frac{B_o (-R \cos 	heta)}{2R} \hat k = -\frac{B_o \cos 	heta}{2} \hat k$.Now, $d\vec F = I [R d	heta (\cos 	heta \hat i + \sin 	heta \hat j)] 	imes [-\frac{B_o \cos 	heta}{2} \hat k]$.$d\vec F = -\frac{I B_o R}{2} \cos 	heta d	heta [\cos 	heta (\hat i 	imes \hat k) + \sin 	heta (\hat j 	imes \hat k)]$.Since $\hat i 	imes \hat k = -\hat j$ and $\hat j 	imes \hat k = \hat i$, we have:$d\vec F = -\frac{I B_o R}{2} [-\cos^2 	heta \hat j + \sin 	heta \cos 	heta \hat i] d	heta$.Integrating from $	heta = 0$ to $\pi$:$F_x = -\frac{I B_o R}{2} \int_0^{\pi} \sin 	heta \cos 	heta d	heta = -\frac{I B_o R}{4} \int_0^{\pi} \sin(2	heta) d	heta = 0$.$F_y = \frac{I B_o R}{2} \int_0^{\pi} \cos^2 	heta d	heta = \frac{I B_o R}{2} \int_0^{\pi} \frac{1 + \cos(2	heta)}{2} d	heta = \frac{I B_o R}{4} [	heta + \frac{\sin 2	heta}{2}]_0^{\pi} = \frac{I B_o R \pi}{4}$.Since the force is in the positive $y$-direction, the correct option is $B$.

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